Need help with Trigo: Inclined Plane and Force

[dandirom]

A 230-Newton box rests on a plank 5 meters long. One end is 1.20 meters higher than the other end.

Find the components of the force the box exerts:

a) parallel to the inclined plane
b) perpendicular to the inclined plane

 

[Hootenanny]

Could you please post what you have attempted thus far?

 

[Architeuthis Dux]

Sure, I'd be happy to help with Trigo and inclined planes! First, let's start by drawing a diagram of the situation to better visualize it. We have a 230-Newton box resting on a plank that is 5 meters long and inclined at an angle of 1.20 meters higher on one end than the other. The box exerts a force on the plank in both a parallel and perpendicular direction.

To find the component of the force parallel to the inclined plane, we can use the formula Fparallel = mg sinθ, where m is the mass of the box, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of inclination. In this case, we can use the given information to calculate the force parallel to the inclined plane as:

Fparallel = (230 N)(9.8 m/s^2) sin(1.20 m/5 m) = 43.29 N

This means that the box exerts a force of 43.29 N parallel to the inclined plane.

To find the component of the force perpendicular to the inclined plane, we can use the formula Fperpendicular = mg cosθ. In this case, the force perpendicular to the inclined plane would be equal to the weight of the box, which is 230 N.

Therefore, the components of the force that the box exerts on the inclined plane are:

a) parallel to the inclined plane = 43.29 N
b) perpendicular to the inclined plane = 230 N

I hope this helps you with your problem! If you have any further questions, please don't hesitate to ask. Best of luck with your Trigo studies!